Answer:
-v/2
Explanation:
Given that:
- Collides with the wall going through a sliding motion on on the plane smooth surface.
- Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.
<u>We know, kinetic energy is given as:</u>
consider this to be the initial kinetic energy of the body.
<u>Now after collision:</u>
Considering that the mass of the body remains constant before and after collision.
Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.
Answer: (1) Putrefaction
(2) Biodegradable
(3) Refuse.
Putrefaction refers to <u>decomposition of organic substances</u> usually dead plants' and animals' bodies <u>by microorganisms</u>.
Therefore, it matches with first definition that states the decay of organic matter caused by microorganisms.
Biodegradable substances are <u>decomposed by living organisms such as microorganisms.</u>
Therefore, it matches with the second definition that states susceptible to decomposition by living organisms.
Refuse refers to something which is <u>no longer useful</u>.
Therefore, it matches with the third definition of waste.
P= 10m = 10 x 180 = 1800 N
work is done by the weightlifter in lifting the barbells :
A= Ph = 1800 x 1.95 = 3510 J
ok done. Thank to me :>
Answer:
Explanation:
The temperature in stratosphere is generally about 270 K
molecular weight of an ozone molecule = 48 gm/mole
now formula for most probable velocity
plugging the values we get
Answer:
the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
Explanation:
Given the data in the question;
first we compute the total volume of mud
V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m
V = 2.508 × 10⁹ m³
same volume of mud spread uniformly over 1.4 × 1.4 km² area
Hence,
V = (1.4×1000)m × (1.4×1000)m × depth
2.508 × 10⁹ = 1400m × 1400m × depth
2.508 × 10⁹ m³ = 1960000m² × depth
depth = 2.508 × 10⁹ m³ / 1960000 m²
depth = 1279.59 m
so volume of the mud sitting above 3.5 m² area of the valley floor will be;
⇒ 3.5 m² × 1279.59 m
⇒ 4478.565 m³
so mass will be
m = 4478.565 × 1900 kg
m = 8509273.5 kg
Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg